Check work on 2 variable function.

[tnutty]

Homework Statement

Determine the set of points at which the function is continuous.

F(x,y) = R.

where R is a piecewise function of :

{
x^2*y^3 / (2x^2 + y^2) ; if(x,y) != (0,0)
1 ; if(x,y) = (0,0)
}

Obviously, the first function is not defined at point (0,0), but to find
the domain of the piecewise function, I first need to see if the
first function is at least continuous.

So here is my attempt at that :

Let A = x^2 * y^3 / (2x^2 + y^2 )

the |A| is =

x^2 * |y^3|
---------------
2x^2 + y^2

well x^2 <= 2x^2 + y^2, let's call that J

so A < J * |y^3| / (J) = |y^3| = sqrt(y^6), and we see that this function
is defined at point 0 , thus lim of A as (x,y) -->(0,0) = 0. ?

So if the above is true then the peicewise function should be
defined in region R^2?

I am not sure if this is correct. The book says that the answer is :

{ (x,y) | (x,y) != (0,0) }.

I think that means the function A is not defined at 0 thus the peicewise
function is not defined at point (0,0). What did I do wrong ?

 

[Billy Bob]

You showed \lim_{(x,y)\to (0,0)} F(x,y)=0 . However, since F(0,0)=1 by the definition of F, we have that F is not continuous at (0,0).

 

[tnutty]

You showed \lim_{(x,y)\to (0,0)} F(x,y)=0 . However, since F(0,0)=1 by the definition of F, we have that F is not continuous at (0,0).

I thought that given value defined in the piecewise, 1 in this case, was an arbitrary value.
So it did not matter, unless the limit of part A in the piecewise function did not exist?

 

[Billy Bob]

If F(0,0) had been defined to be 0 instead of 1, then F would have been continuous everywhere.

If F(0,0) had been defined to be c, with c nonzero (c=1 is a special case), then as in your problem, F would not have been continuous at (0,0) but would be continuous everywhere else.